CONCEPT:

Change in Internal Energy of a Gas

• To determine the change in internal energy (ΔU), we use the first law of thermodynamics:
• ΔU = Q - W
• Where:
  • ΔU: Change in internal energy
  • Q: Heat added to the system (negative if heat is removed)
  • W: Work done by the system
• For a gas compressed or expanded at constant pressure, the work done (W) is given by:
• W = P ΔV
• Where:
  • P: Constant pressure
  • ΔV: Change in volume

CALCULATION:

• Given:
  • Initial Volume (V₁) = 325 ml = 0.325 L
  • Final Volume (V₂) = 125 ml = 0.125 L
  • Constant Pressure (P) = 2 bar = 2 × 10⁵ Pa
  • Heat removed (Q) = -124 J
• First, calculating the change in volume (ΔV):
  • ΔV = V₂ - V₁ = 0.125 L - 0.325 L = -0.2 L
  • Convert the change in volume to cubic meters (m³):
    • ΔV = -0.2 L × 10^-3 m³/L = -0.0002 m³
• Now, work done (W):
  • W = P ΔV = (2 × 10⁵ Pa) × (-0.0002 m³) = -40 J
• Appling the first law of thermodynamics to find the change in internal energy (ΔU):
  • ΔU = Q - W = -124 J - (-40 J) = -124 J + 40 J = -84 J

CONCLUSION:

The change in internal energy when the gas contracts from 325 ml to 125 ml at a constant pressure of 2 bar, while removing 124 J of heat, is:

ΔU = -84 J

CONCEPT:

Entropy and Heat Capacity Relationship

• The relationship between heat capacity (C) and entropy (S) is given by:

• Where S₀ is the constant of integration, which can be considered as the entropy at reference temperature (usually at absolute zero, S₀ = 0).

CONCLUSION:

The entropy (S) of a crystal at very low temperatures, where the heat capacity is given by 

Concept:

To determine the enthalpy of dehydration of blue vitriol (Copper (II) sulphate pentahydrate), we use the enthalpy changes of related processes:

• Enthalpy of Solution (ΔH_sol): The change in enthalpy when one mole of a substance dissolves in a solvent.
  • For anhydrous CuSO₄: ΔH_sol = -66.11 kJ/mol
  • For CuSO₄•5H₂O: ΔH_sol = +11.5 kJ/mol (since heat is absorbed)
• Enthalpy of Dehydration: The enthalpy change when water is removed from a hydrated compound.
• Hess's Law: The total enthalpy change of a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.

Explanation:

1. Step 1 - Dissolution of CuSO₄•5H₂O:
   • CuSO₄•5H₂O (s) → Cu²⁺ (aq) + SO₄^2- (aq)
   • ΔH_{sol, CuSO4•5H2O} = +11.5 kJ/mol
2. Step 2 - Dissolution of anhydrous CuSO₄:
   • CuSO₄ (s) → Cu²⁺ (aq) + SO₄^2- (aq)
   • ΔH_{sol, CuSO4} = -66.11 kJ/mol
3. Step 3 - Subtracting the two processes to find the enthalpy of dehydration:
   • ΔH_dehydration = ΔH_{sol, CuSO4•5H2O} - ΔH_{sol, CuSO4}
   • ΔH_dehydration = 11.5 kJ/mol - (-66.11 kJ/mol)
   • ΔH_dehydration = 11.5 kJ/mol + 66.11 kJ/mol
   • ΔH_dehydration = 77.61 kJ/mol

Conclusion:

Therefore, the enthalpy of dehydration of blue vitriol (Copper (II) sulphate pentahydrate) is +77.61 kJ/mol. Hence, the correct answer is: +77.61 kJ

CONCEPT:

1) Reduction Reaction:

• Involves the gain of electrons (decrease in oxidation state).
• Example: Fe³⁺ + e^- → Fe²⁺

2) Disproportionation Reaction:

• Occurs when a single substance is simultaneously oxidized and reduced.
• Example: 2H₂O₂ → 2H₂O + O₂

3) Oxidation Reaction:

• Involves the loss of electrons (increase in oxidation state).
• Example: Fe²⁺ → Fe³⁺ + e^-

4) Decomposition Reaction:

• A single compound breaks down into two or more simpler substances.
• Example: 2H₂O → 2H₂ + O₂

EXPLANATION:

The given reaction is:

3 ClO^- → ClO₃^- + 2 Cl^-

• Oxidation States:
  • For ClO^-: The oxidation state of Cl is +1.
  • For ClO₃^-: The oxidation state of Cl is +5.
  • For Cl^-: The oxidation state of Cl is -1.
• Oxidation and Reduction:
  • Cl in ClO^- changes from +1 to +5 in ClO₃^- (oxidation).
  • Cl in ClO^- changes from +1 to -1 in Cl^- (reduction).
• This reaction involves the same species (ClO^-) being both oxidized and reduced.

CONCLUSION:

This reaction is an example of: Disproportionation reaction

CONCEPT:

General Rules for IUPAC Nomenclature:

• Identify the longest continuous carbon chain as the parent hydrocarbon.
• Number the carbon atoms in the parent chain starting from the end nearest a substituent.
• Name and number the substituents. Prefix the names of substituents with the carbon number to which they are attached.
• When more than one substituent is present, list them in alphabetical order.
• If two or more identical substituents are present, use prefixes such as di-, tri-, tetra-, etc., but these prefixes are not included in the alphabetical order.
• Use appropriate suffixes for functional groups (e.g., -ol for alcohols, -one for ketones, etc.).

• Identify the longest continuous chain: Hexane (6 carbon atoms).
• Substituents and their positions:
  • Hydroxy group at carbon 2.
  • Hydroxymethyl group at carbon 3.
  • Hydroxy and methyl group at carbon 4.

Thus, the correct name of the compound is 3-hydroxy methyl-4-methyl hexane-2,4-diol.

CONCLUSION:

The correct IUPAC name of the given compound is: 3-hydroxy methyl-4-methyl hexane-2,4-diol

Concept:

General Rules of IUPAC Nomenclature and Priority of Functional Groups

• Identify the Principal Functional Group: The functional group with the highest priority is chosen as the suffix of the compound name. Common functional groups in order of priority include -COOH (carboxylic acid), -OH (alcohol), -CO (keto), -NH₂ (amine), and others.

• Identify the Longest Chain Including the Functional Group: Find and name the longest continuous carbon chain that includes the highest priority functional group.

• Number the Carbon Chain: Number the chain such that the principal functional group gets the lowest possible number.

• Identify and Number Substituents: Identify all substituents and assign numbers to them based on their position on the carbon chain.

• Combine the Elements: Combine the numerical locants, substituent names, and the parent hydrocarbon name with the appropriate suffix for the principal functional group.

• Use of Prefixes for Identifying Substituents: Use appropriate prefixes for identifying other substituents such as alkyl, acyl, and aryl groups.

Explanation:

To identify the IUPAC name for the compound with the given structure:

• The principal functional group is the carboxylic acid (-COOH), which has the highest priority.

• The longest carbon chain that includes the carboxylic acid is a three-carbon chain, hence the parent name "propanoic acid."

• The benzoyloxy group (C₆H₅-CO-O-) is attached to the third carbon of the propanoic acid chain.

Conclusion:

The IUPAC name of the compound is: 3-(Benzoyloxy)propanoic acid

Concept:

Isomers in Organic Chemistry

Isomers are compounds that have the same molecular formula but different structures or spatial arrangements of atoms. Here are key points and different types of isomers:

Types of Isomers

1. Structural Isomers (Constitutional Isomers)

• Chain Isomers: Compounds with different carbon chain arrangements. Example: n-butane and isobutane.

• Position Isomers: Compounds with the same carbon skeleton and functional groups but different positions of the functional group. Example: 1-butanol and 2-butanol.

• Functional Group Isomers: Compounds with the same atoms arranged into different functional groups. Example: ethanol and dimethyl ether.

2. Stereoisomers

• Geometric Isomers (cis-trans Isomers): Compounds with restricted rotation around a bond, typically a double bond or ring structure, resulting in different spatial arrangements. Example: cis-2-butene and trans-2-butene.

• Optical Isomers (Enantiomers and Diastereomers): Compounds that differ in the arrangement of atoms in three-dimensional space. Enantiomers are non-superimposable mirror images (e.g., left-hand and right-hand forms of chiral molecules), while diastereomers are not mirror images. Example: (R)- and (S)-lactic acid.

​Explanation:

The arrangement of parent chain is different. So, they become chain isomer.

Concept:

Thus, pair of chain isomers from the given compounds are: II and III.

Concept:

Keto-Enol Tautomerism

Keto-enol tautomerism is a chemical equilibrium between a ketone (or aldehyde) and its corresponding enol form (a compound with a hydroxyl group attached to a double-bonded carbon). 

Factors Affecting the Extent of Keto-Enol Tautomerism:

• Nature of the Compound: Simple ketones and aldehydes usually favor the keto form due to the stability of the carbonyl group. However, compounds with electron-withdrawing groups or conjugation can stabilize the enol form.

• Solvent: Polar solvents, especially those capable of hydrogen bonding (like water), often stabilize the enol form, while non-polar solvents may favor the keto form.Temperature: Higher temperatures can favor the enol form due to its potential higher entropy.

• Acidity of the Environment: Acidic conditions can promote enolization by protonating the keto form, whereas basic conditions can deprotonate the alpha hydrogen, facilitating enolate formation.a

• Presence of Hydrogen Bonding: Intramolecular hydrogen bonds can stabilize the enol form in certain compounds, shifting the equilibrium towards the enol form.

• Structure and Substituents: Conjugation with double bonds or aromatic rings can stabilize the enol form through resonance. Additionally, alpha-substituents that withdraw electrons (like nitro or carbonyl groups) may stabilize the enol form.

​Explanation:

This enol form is stabilized by Hydrogen bonding and also, the additional stabilization due to resonance of Phenyl group. 

Conclusion:

 has the highest enol content.

Concept:

The stability of anions is influenced by several factors such as inductive effects, resonance, hyperconjugation, and the nature of the substituents. Key points include:

• Inductive Effect: Electron-withdrawing groups stabilize anions through the inductive effect.

• Resonance Stabilization: Anions that can delocalize their negative charge through resonance are more stable.

• Hyperconjugation: Hyperconjugation can stabilize anion through delocalization of electrons.

• Alkyl Substituents: More alkyl substituents can destabilize an anion due to their electron-donating nature.

Explanation:

• (CH₃)₃C^– (I): Tertiary alkyl anion. Least stable due to the strong electron-donating nature of the three alkyl groups.

• (CH₃)₂CH^– (II): Secondary alkyl anion. Less stable than primary alkyl anions.

• CH₃CH₂^– (III): Primary alkyl anion. More stable than secondary and tertiary due to fewer electron-donating groups, but not as stable as benzyl anion.

• C₆H₅CH₂^– (IV): Benzyl anion. Highly stabilized due to resonance with the aromatic ring, making it the most stable among these anions.

Conclusion:

Therefore, the correct answer is: IV > III > II > I

Concept:

The basicity of a compound is determined by the availability of the lone pair of electrons on the nitrogen atom for protonation. Factors affecting basicity include hybridization, electron delocalization, and the molecular structure of the compound. Below are some key points:

• Hybridization: sp² hybridized nitrogen is less basic than sp³ hybridized nitrogen because the lone pair is held more tightly.

• Aromatic Character: An aromatic system can influence the lone pair availability on the nitrogen.

• Cyclic Structure: Cyclic systems can impact electronic distribution and lone pair availability.

• Lone Pair Delocalization: If the lone pair is delocalized, it is less available for protonation, decreasing basicity.

Explanation:

Pyridine (C₅H₅N) is compared to triethylamine (N(CH₂CH₃)₃) to understand basicity differences:

• Pyridine: Contains a nitrogen atom in an aromatic six-membered ring.

  • Hybridization: Nitrogen in pyridine is sp² hybridized.

  • Lone Pair Delocalization: The lone pair on the nitrogen is part of the aromatic system and is delocalized, making it less available for protonation resulting in less basicity.

• Triethylamine: Contains a nitrogen atom bonded to three ethyl groups.

  • Hybridization: Nitrogen in triethylamine is sp³ hybridized.

  • Lone Pair Availability: The lone pair is localized and more available for protonation, increasing basicity.

Conclusion:

The key reason that pyridine is less basic than triethylamine is due to: Lone pair of nitrogen is delocalised in pyridine